Question

Evaluate the following integrals : $$\int \frac{\left(x^{2}-1\right) d x}{x \sqrt{\left(x^{2}+a x+1\right)\left(x^{2}+b x+1\right)}}$$

Short answer

In the given integral, the most important step in solving it is to simplify the expression under the square root using substitution methods. We first substituted variables for the quadratic expressions and then rewrote the integral in terms of the new variables. We then simplified the integral expression and found the antiderivatives. The final step was converting back to the original variable, x. The solution to the definite integral is: $$\int \frac{\left(x^{2}-1\right) d x}{x \sqrt{\left(x^{2}+a x+1\right)\left(x^{2}+b x+1\right)}} = \ln\left(\sqrt{\left((x + \frac{a - 1}{2})^{2}+1\right)\left((x + \frac{b - 1}{2})^{2}+1\right)}\right) - \frac{(x-a)(x+a)}{4} \ln(x + \frac{a - 1}{2}) + C$$

Step-by-step solution

Substitute variables for quadratic expressions

To perform substitution, let's first consider the quadratic expressions in the denominator: $$f(x) = x^2 + ax + 1$$ $$g(x) = x^2 + bx + 1$$ Let us substitute \(x = u + \frac{a - 1}{2}\) for \(f(x)\) and \(x = u + \frac{b - 1}{2}\) for \(g(x)\). Then we have: $$f(u) = u^2 + 1$$ $$g(v) = v^2 + 1$$ Now, we'll express our integral in terms of \(u\) and \(v\).

Rewrite the integral in terms of \(u\) and \(v\)

Substitute \(x\) in terms of \(u\) and \(v\) in the integral: $$\int \frac{\left(\left(u + \frac{a - 1}{2}\right)^{2}-1\right) du}{\left(u + \frac{a - 1}{2}\right) \sqrt{\left(u^2+1\right)\left(v^{2}+1\right)}}$$ Observe that the expression in the numerator becomes: $$\left(u^2 + \frac{(a-1)(a+1)}{4}\right)-1 = u^2 + \frac{a^2-1}{4}$$ So, the integral can be simplified to: $$\int \frac{\left(u^{2} + \frac{a^{2}-1}{4}\right) du}{u \sqrt{\left(u^2+1\right)\left(v^{2}+1\right)}}$$

Simplify the integral expression

To simplify the integral, let's split the numerator into two terms, and then deal with each term individually: $$\int \frac{u^2 du}{u \sqrt{\left(u^2+1\right)\left(v^{2}+1\right)}} + \int \frac{\frac{a^{2}-1}{4} du}{u \sqrt{\left(u^2+1\right)\left(v^{2}+1\right)}}$$ Now, we can further simplify each term: $$\int \frac{u du}{\sqrt{(u^2+1)(v^2+1)}} + \frac{a^{2}-1}{4} \int \frac{du}{u\sqrt{(u^2+1)(v^2+1)}}$$

Find antiderivatives

Now, we only need to compute the antiderivatives: $$\int \frac{u du}{\sqrt{(u^2+1)(v^2+1)}} = \ln(\sqrt{(u^2+1)(v^2+1)})$$ and $$\frac{a^{2}-1}{4} \int \frac{du}{u\sqrt{(u^2+1)(v^2+1)}} = \frac{a^{2}-1}{4} \ln(u)$$ So, the solution to the integral is: $$\ln\left(\sqrt{(u^2+1)(v^2+1)}\right) - \frac{a^{2}-1}{4} \ln(u) + C$$

Convert back to \(x\)

Finally, let's convert back to \(x\): $$\ln\left(\sqrt{\left((x + \frac{a - 1}{2})^{2}+1\right)\left((x + \frac{b - 1}{2})^{2}+1\right)}\right) - \frac{(x-a)(x+a)}{4} \ln(x + \frac{a - 1}{2}) + C$$ So, the solution to the given integral is: $$\int \frac{\left(x^{2}-1\right) d x}{x \sqrt{\left(x^{2}+a x+1\right)\left(x^{2}+b x+1\right)}} = \ln\left(\sqrt{\left((x + \frac{a - 1}{2})^{2}+1\right)\left((x + \frac{b - 1}{2})^{2}+1\right)}\right) - \frac{(x-a)(x+a)}{4} \ln(x + \frac{a - 1}{2}) + C$$

Key concepts

Integration Techniques

When tackling mathematical problems, especially integral calculus, understanding various integration techniques becomes fundamental. These techniques are the tools necessary to solve complex integrals that cannot be solved by a straightforward application of basic integration rules.

Consider our example, where the integral involves a complex fraction with polynomials under a square root. Such integrals usually require a little more finesse than simply finding an antiderivative. A variety of techniques may be used depending on the nature of the function being integrated, such as integration by parts, trigonometric substitution, partial fractions, or the substitution method.

The substitution method, in particular, is a powerful tool that simplifies integrals by transforming them into a form that is easier to integrate. The exercise demonstrates how substitution can effectively simplify the problem by reducing the complexity of the denominator and transforming the function into one that involves basic functions whose antiderivatives are known.

Substitution Method

The substitution method is an integration technique that aims to simplify the integrand by replacing it with another variable. This process can dramatically reduce the complexity of an integral, making it easier to evaluate. Essentially, you choose a new variable to substitute into the integral that will simplify the expression, and then find the differential of that new variable so you can replace the dx (or equivalent variable of integration) in the integral as well.

In our example, we replaced the complex quadratic expressions by a single variable squared plus one. This strategic choice in substitution meant that the resulting integral was far simpler to evaluate. We then needed to express the original variable x in terms of the new variables u and v and substitute it throughout the integral. Importantly, we calculated the new differential du to replace dx.

It’s crucial to remember that once the integral is solved in the new variable, the last step is to revert back to the original variable to complete the solution. In some cases, the antiderivative may involve the natural logarithm, as seen in the given solution.

Definite Integrals

While the exercise example doesn't directly deal with definite integrals, it’s important to understand their role in calculus. A definite integral has upper and lower limits on the integral sign, and the result is a number representing the area under the curve of the integrand between these limits.

The process of finding a definite integral is two-fold: first, find the indefinite integral, which includes the antiderivative plus a constant of integration, and then evaluate this antiderivative at the upper and lower limits, subtracting the latter from the former. This is known as the Fundamental Theorem of Calculus.

In cases where the integrand is complex and requires substitution, it's necessary to also change the limits of integration to match the substitution variable. However, one can also solve the indefinite integral first and then revert back to the original variable before evaluating between the limits, just as in our exercise example.

Definite integrals are used across all fields of science and engineering to determine quantities such as areas, volumes, and other sizes that can be defined as sums of infinitely many infinitesimally small parts.